There are no restrictions on zeros of such functions $Lf$ (unless you somehow restrict your class $C$ of functions $f$). For example, let $f$ be an infinite sum of $\delta$-functions sitting at integers. Then your integral becomes a sum, and changing the variable to $z=e^{-s}$ we obtain an arbitrary Laurent series in $\mathbf{C}\backslash\{0\}$. This can have arbitrary sequence of zeros.

The main condition of Lerch's theorem is that $f(x)=0$ for $x<0$, which ensures that $LF$ is bounded in the right half-plane. If we allow an arbitrary support of $f$, no conclusion about zeros can be made, except, of course that this is a discrete set.

Of course, if $A$ has accumulation points then your statement is correct, since you assume the function $Lf$ to be entire.

Otherwise, there are no restrictions on zeros of such functions $Lf$ (unless you somehow restrict your class $C$ of functions $f$). For example, let $f$ be an infinite sum of $\delta$-functions sitting at integers. Then your integral becomes a sum, and changing the variable to $z=e^{-s}$ we obtain an arbitrary Laurent series in $\mathbf{C}\backslash\{0\}$. This can have arbitrary sequence of zeros.

Of course a sum of delta-functions is not integrable, but it is easy to modify this example, to make it integrable by integrating by prts: $$\sum_{-\infty}^\infty a_ke^{-sk}=\int_{-\infty}^\infty e^{-sx}dn(x)=-s\int_{-\infty}^\infty n(x)e^{-sx}dx,$$ where $n$ is a step function jumping by $a_k$ at $k$. The integral in the RHS have the same zeros as the LHS, except at $s=0$, and $n(x)$ is an integrable (step) function. By integrating few more times, you can make your $f$ arbitrarily smooth.

The main condition of Lerch's theorem is that $f(x)=0$ for $x<0$, which ensures that $LF$ is bounded in the right half-plane. If we allow an arbitrary support of $f$, no conclusion about zeros can be made, except, of course that this is a discrete set.